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我使用 ZLayer 和 Sttp Client(async 来创建简单的 http 请求者应用程序，但我发现我无法解决的类型不匹配错误。谁能告诉我为什么我收到类型不匹配错误？

我使用这些版本的 scala 和库。

java -> 8.282.08.1-amzn

斯卡拉-> s.13.5

dev.zio -> 1.0.7

com.softwaremill.sttp.client3 -> 3.3.0

type mismatch;
 found   : zio.ZLayer[sttp.client3.asynchttpclient.zio.SttpClient,Nothing,ZlayerAndSttp.HttpBin]
    (which expands to)  zio.ZLayer[zio.Has[sttp.client3.SttpBackend[zio.Task,sttp.capabilities.zio.ZioStreams with sttp.capabilities.WebSockets]],Nothing,zio.Has[ZlayerAndSttp.HttpBin.Service]]
 required: zio.ZLayer[ZlayerAndSttp.HttpBin,?,?]
    (which expands to)  zio.ZLayer[zio.Has[ZlayerAndSttp.HttpBin.Service],?,?]
    program.provideCustomLayer((AsyncHttpClientZioBackend.layer() >>> HttpBin.live) >>> HttpBin.live)

这是整个代码

import zio._
import sttp.client3._
import sttp.client3.circe._
import sttp.client3.asynchttpclient.zio._
import io.circe.generic.auto._
import zio.console.Console


object ZlayerAndSttp extends App {

  case class HttpBinResponse(origin: String, headers: Map[String, String])

  type HttpBin = Has[HttpBin.Service]
  object HttpBin {
    trait Service {
      def sendRequest: ZIO[HttpBin with SttpClient, Throwable, HttpBinResponse]
    }

    val live: ZLayer[SttpClient, Nothing, HttpBin] = ZLayer.succeed(new Service {
      override def sendRequest: ZIO[HttpBin with SttpClient, Throwable, HttpBinResponse] = {
        val request = basicRequest
          .get(uri"https://httpbin.org/get")
          .response(asJson[HttpBinResponse])
        sendR(request).map(_.body).absolve.map(res => HttpBinResponse(res.origin, res.headers))
      }
    })

    def sendRequest: ZIO[HttpBin with SttpClient, Throwable, HttpBinResponse] = ZIO.accessM(_.get.sendRequest)
  }

  val request = basicRequest
    .get(uri"https://httpbin.org/get")
    .response(asJson[HttpBinResponse])

  override def run(args: List[String]): URIO[zio.ZEnv, ExitCode] = {
    val program = for {
      result <- HttpBin.sendRequest
      _ <- console.putStrLn(s"${result.origin}, ${result.headers}")
    } yield ()
    program.provideCustomLayer((AsyncHttpClientZioBackend.layer() >>> HttpBin.live) >>> HttpBin.live) // type mismatch
      .exitCode

    // ↓these lines of code run with no errors but I can't understand why
//    val program: ZIO[Console with SttpClient, Throwable, Unit] = for {
//      response <- send(request)
//      _ <- console.putStrLn(s"${response.body.toString}")
//    } yield ()
//    program.provideCustomLayer(AsyncHttpClientZioBackend.layer()).exitCode
  }

}
4

1 回答 1

0

你似乎让你的生活变得比需要的更复杂一些。

最终代码的外观取决于您要在此处实现的目标。如果您试图隐藏界面Sttp后面的使用HttpBin，那么您的层定义应该如下所示：

    val live: ZLayer[SttpClient, Nothing, HttpBin] = 
      (for {
        client <- ZIO.environment[SttpClient]
      } yield new Service {
          override def sendRequest: ZIO[Any, Throwable, HttpBinResponse] = {
            val request = basicRequest
              .get(uri"https://httpbin.org/get")
              .response(asJson[HttpBinResponse])

            sendR(request)
              .map(_.body)
              .absolve
              .map(res => HttpBinResponse(res.origin, res.headers))
              .provide(client)
  }
      }).toLayer

然后您的访问器方法变为：

def sendRequest: ZIO[HttpBin, Throwable, HttpBinResponse] = 
  ZIO.accessM(_.get.sendRequest)

你可以使用它：

  override def run(args: List[String]): URIO[zio.ZEnv, ExitCode] = {
    val program = for {
      result <- HttpBin.sendRequest
      _ <- console.putStrLn(s"${result.origin}, ${result.headers}")
    } yield ()

    program
      .provideCustomLayer(AsyncHttpClientZioBackend.layer() >>> HttpBin.live)
      .exitCode

这里要注意的是，对于图层组合，我只使用垂直运算符，因为它HttpBin.live依赖于一个图层，SttpClient但我们从调用方法中“隐藏”了这一事实，因此如果需要，您可以创建一个不需要 Sttp的test变体HttpBin.

如果您不需要隐藏信息，则可以完全删除中间层，并将您的sendRequest作为独立方法。

object HttpBin {
  def sendRequest: ZIO[SttpClient, Throwable, HttpBinResponse] = {
     val request = basicRequest
       .get(uri"https://httpbin.org/get")
       .response(asJson[HttpBinResponse])

     sendR(request)
       .map(_.body)
       .absolve
       .map(res => HttpBinResponse(res.origin, res.headers))
}

然后你可以调用这个方法，你需要做的就是提供SttpClient层。

于 2021-05-09T13:16:19.740 回答